The diameter and altitude of a right circular | vedclass

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(D) Let $D$ be the diameter,$r$ be the radius,and $h$ be the altitude of the right circular cone. Given $D = 10 \ cm$,so $r = 5 \ cm$. Given $h = 20 \

Vedclass · Accepted Answer

$-8$

Vedclass · Answer

(D) Let $D$ be the diameter,$r$ be the radius,and $h$ be the altitude of the right circular cone. Given $D = 10 \ cm$,so $r = 5 \ cm$. Given $h = 20 \ cm$. The rate of change of diameter is $\frac{dD}{dt} = 2 \ cm/s$. Since $D = 2r$,we have $\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt} = \frac{1}{2} 	imes 2 = 1 \ cm/s$. The volume of the cone is $V = \frac{1}{3} \pi r^2 h$. To keep the volume constant,$\frac{dV}{dt} = 0$. Using the product rule,$\frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$. Substituting the values: $2(5)(20)(1) + (5)^2 \frac{dh}{dt} = 0$. $200 + 25 \frac{dh}{dt} = 0$. $25 \frac{dh}{dt} = -200$. $\frac{dh}{dt} = -8 \ cm/s$. Thus,the altitude must change at a rate of $-8 \ cm/s$.

The diameter and altitude of a right circular cone,at a certain instant,were found to be $10 \ cm$ and $20 \ cm$ respectively. If its diameter is increasing at a rate of $2 \ cm/s$,then at what rate must its altitude change,in order to keep its volume constant (in $cm/s$)?

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